What is the integral of (3-x)^3 ?

The integral of (3-x)^3 is -((3-x)^4)/4+C

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integral of (3-x)^3

\int (3 - x)^{3} d x

- \frac{( 3 - x ) ^{4}}{4} + C

Solution steps

\int (3 - x)^{3} d x

Apply u-substitution

: \int - u^{3} d u

= \int - u^{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{u ^{4}}{4}

Substitute back

u = 3 - x

= - \frac{( 3 - x ) ^{4}}{4}

Add a constant to the solution

= - \frac{( 3 - x ) ^{4}}{4} + C

y = x^{4} + 8 e^{x}, (0, 8)

\int cos (2 x + 3) d x

\frac{d}{d x} (4 - x^{3})

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}, y (0) = 7

x \to 7 - lim (∣ x - 7 ∣)